Evaluate: 2 x 5 x 3 x d x - Vidyadip

Question

Evaluate: $\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x$

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Answer

Let $I =\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x$.
Then, we have
$I=\int \frac{\sin (5 x-3 x)}{\sin 5 x \sin 3 x} d x$
$=\int \frac{\sin 5 x \cos 3 x-\cos 5 x \sin 3 x}{\sin 5 x \sin 3 x} d x$
$=\int \frac{\sin 5 x \cos 3 x}{\sin 5 x \sin 3 x} d x-\int \frac{\cos 5 x \sin 3 x}{\sin 5 x \sin 3 x} d x$
$=\int \frac{\cos 3 x}{\sin 3 x} d x-\int \frac{\cos 5 x}{\sin 5 x} d x$
$=\int \cot 3 x d x-\int \cot 5 x d x$
$=\frac{1}{3} \log |\sin 3 x|-\frac{1}{5} \log |\sin 5 x|+c$
$\therefore I=\frac{1}{3} \log |\sin 3 x|-\frac{1}{5} \log |\sin 5 x|+c$

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